Step 1.  Translate line by line into a picture, as in Figure 1, and math.

A picture doesn’t make sense for this problem.  So, we’ll have to jump straight into translating the problem into math.  We are trying to maximize revenue, and we know that [latex]R=pq[/latex], where [latex]p[/latex] is the price per ticket, and [latex]q[/latex] is the quantity of tickets sold.

The problem provides information about the demand relationship between price and quantity – as price increases, demand decreases. We need to find a formula for this relationship. To investigate, let’s calculate what will happen to attendance if we raise the price, organizing our thoughts in Table 1:

Table 1

You might recognize this as a linear relationship. We can find the slope for the relationship by using two points: [latex]m=\frac{1150-1200}{55-50}=\frac{-50}{5}=-10.[/latex]

You may notice that the second step in that calculation corresponds directly to the statement of the problem: the attendance drops 50 people for every $5 the price increases.

Using the point-slope form of the line, we can write the equation relating price and quantity: [latex]q-1200=-10(p-50).[/latex]

Simplifying to slope-intercept form gives the demand equation [latex]q=1700-10p.[/latex]

Substituting this into our revenue equation, we get an equation for revenue involving only one variable: [latex]R=pq=p(1700-10p)=1700p-10p^2.[/latex]

Step 2.  Identify the objective function.

The “buzz” word in this problem that helps us identify the objective function is “maximize.”  The problem asked us to “maximize revenue.”  A maximum is something we know we can use calculus to find in Step 3.  So,

[latex]\begin{array}{llll}R \hfill & = & pq & \\ R \hfill &= & p(1700-10p) \hfill & \\ R \hfill & = & 1700p-10{p}^{2} \hfill & \text{is our objective function of one variable.} \hfill \end{array}[/latex]

Step 3.  Use calculus to find the optimum values.

Now, we can find the maximum of this function by finding critical numbers. [latex]R'=1700-20p[/latex], so [latex]R'=0[/latex] when [latex]p = 85[/latex].

Using the second derivative test, [latex]R''=-20 \lt 0[/latex] (for any value of [latex]p[/latex]), so the critical number is a local maximum. Since it is the only critical number, we can also conclude that it is the global maximum.

The promoter will be able to maximize revenue by charging $85 per ticket. At this price, she will sell [latex]q=1700-10(85)=850[/latex] tickets, generating $72,250 in revenue.

Step 4.  Look back at the question to make sure you answered what was asked.

The question asked us to find the price that would maximize revenue.  To that end, we

• gave an answer of \$85, which represents a ticket’s sale price,  [latex]\checkmark[/latex]
• used the 2nd derivative to guarantee that this price results in the maximum revenue,  [latex]\checkmark[/latex]
• and used common sense to judge whether this answer makes sense in the context of the problem.  [latex]\checkmark[/latex]

Step 1.  Translate line by line into a picture, as in Figure 1, and math.

A picture doesn’t make sense for this problem either.  So, like Example 2, we’ll jump straight into translating the problem into math.  We want to maximize profit, but there isn’t a formula for profit given. S o let’s make one.

We can find a function for Revenue = [latex]pq[/latex] using the demand function for [latex]p[/latex]. [latex]R(q)=(300-0.02q)q=300q-0.02q^2.[/latex]

We can also find a function for Cost, using the variable cost of $30 per ribbon winder, plus the fixed cost: [latex]C(q)=9000+30q.[/latex]

Putting them together, we get a function for Profit: [latex]P(q)=R(q)-C(q)=\left(300q-0.02q^2\right)-(9000+30q)=-0.02q^2+270q-9000[/latex]

Step 2.  Identify the objective function.

The “buzz” word in this problem that helps us identify the objective function is “maximized.”  The problem asked us to find where “profit is maximized.”  A maximum is something we know we can use calculus to find in Step 3.  So,

[latex]\begin{array}{llll} P(q) \hfill & = & R(q)-C(q) & \\ P(q) \hfill &= & \left(300q-0.02q^2\right)-(9000+30q) \hfill & \\ P(q) \hfill & = & -0.02q^2+270q-9000 \hfill & \text{is our objective function of one variable.} \hfill \end{array}[/latex]

Step 3.  Use calculus to find the optimum values.

Now we have two choices. We can find the critical points of Profit by taking the derivative of [latex]P(q)[/latex] directly, or we can find [latex]MR[/latex] and [latex]MC[/latex] and set them equal. (Naturally, we’ll get the same answer either way.)

Let’s use [latex]MR = MC[/latex] this time.

[latex]\begin{align*} MR=& 300-0.04q\\ MC=& 30\\ 300-0.04q=& 30\\ 270=& 0.04q\\ q=& 6750 \end{align*}[/latex]

The only critical point is at [latex]q = 6750[/latex]. Now we need to be sure this is a local max and not a local min. In this case, we’ll look to the graph of [latex]P(q)[/latex] – it’s a downward opening parabola, so this must be a local max. And since it’s the only critical point, it must also be the global max.

Profit is maximized when they sell 6750 ribbon winders.

Step 4.  Look back at the question to make sure you answered what was asked.

The question asked us to find the quantity that would maximize profit.  To that end, we

• gave an answer of 6750 ribbon winders, [latex]\checkmark[/latex]
• used the graph of profit to verify that this is a maximum,  [latex]\checkmark[/latex]
• and used common sense to judge whether this answer makes sense in the context of the problem.  [latex]\checkmark[/latex]